Question

If $f\left(x\right)=sinx+cosx$ and $g\left(x\right)=x^{2}-1$ , then $g(f (x))$ is invertible in the domain

• A. $\left[0 , \frac{\pi }{2}\right]$
• B. $\left[- \frac{\pi }{4} , \frac{\pi }{4}\right]$
• C. $\left[- \frac{\pi }{2} , \frac{\pi }{2}\right]$
• D. $\left[0 , \pi \right]$

Answer 1

Answer: (B)

Given, $f\left(x\right)=sinx+cosx$
$g\left(x\right)=x^{2}-1$
$g(f(x)=(f (x))^{2}-1=(sinx + cos x)^{2}-1=sin2x$
But $sin2x$ is invertible in a case when $-\frac{\pi }{2}\leq 2x\leq \frac{\pi }{2}$
⇒ $-\frac{\pi }{4}\leq x\leq \frac{\pi }{4}$